Vehicle Structure Analysis

Asst. Prof. Dr. Kaukeart Boonchukosol

Fundamental Vehicle loads and their estimation

Sampling of thecustomer loadenvironment onpublic road.

Sample of vehicles

Measuring their useacross applicableregion

Update company provingground road schedule.

(accelerated test)

Actual Process

Type of load No. of eventrepetitions

Loadamplitude

(N)

Acceptancecriteria

Typical Proving Ground Events

Instantaneousoverload

Fatigue

Low(< 10)

High(< 102)

High(104)

Low(103)

Cycles to crack initiations,Limited crack propagation,Maintenance of function

Limited permanent deformation,Maintenance of function

1. Ensure the structure will not fail in service.2. Ensure satisfactory fatigue life

Objectives

Load Analysis

If the structure can resist the (rare) worst possible loading which can be encountered, then it is likely to have sufficient fatigue strength.

Dynamic load = Static load x Dynamic load factor

Equivalent load = Static load x Dynamic load factor x Safety factor

Sometimes an extra “factor of safety” is used.

For an early design stage, it is usually assumed that

Basic Global Load Cases

1. Vertical symmetrical (Bending load case)

2. Vertical asymmetrical (Torsion load case)

3. Fore and aft loads (Braking, Acceleration etc.)

4. Lateral loads (Cornering, nudging kerb)

5. Local load cases (Door slam etc.)

6. Crash load

Vertical Symmetric Load Case

Sources :

1. Weight of the major components2. Payload3. Simultaneously bump

Dynamic consideration :

1.4~1.6(away from stres concentration)

1.5~2.0(engine and suspension mounting)

Commonlyused

Pawlowski(1969)

Dynamicfactor

Safetyfactor

3

1.5

2

Erz(1957)

2.5

ha

DA

RF

RR

ab

L

Fxf

Fxr

Rxr

Rxf

(W/g) ax

WsinθWcosθdh

hh

h

RhZ

Rhx

W is the weight of the vehicle

Rxf , Rxr

Fxf , FxrRF , RR

is the rolling resistance force

is the tractive forceis the wheel reaction(dynamic weight)

Rhx , Rhz

DAis the aerodynamic force

W

is the towing force

Consider the vehicle shown below, in which most of significant forceson the vehicle are shown.

[ ]0=Σ AM +

( ) ( ) 0cossin =−+++++ bWhWdRhRhag

WhDLR hhzhhxxaAF θθ

Equilibrium equations

[ ]0=Σ BM +

( ) ( ) ( ) 0cossin =+++++++− aWhWLdRhRhag

WhDLR hhzhhxxaAR θθ

( ) ( ) ⎥⎦

⎤⎢⎣

⎡−−−−−= hWhDha

gWdRhRbW

LR aAxhhzhhxF θθ sincos1

( ) ( ) ( ) ⎥⎦

⎤⎢⎣

⎡++++++= hWhDha

gWLdRhRaW

LR aAxhhzhhxR θθ sincos1

Vertical Asymmetric Load Case

Asymmetric loading is specified by themaximum height H of a bump upon which one wheel of one axle rests, with all other wheels on level ground.

θTKT =

Front axle

Rear axle

H Body

TB

RBFT KKKK1111

++=

BH

≈θ

and

where

KF, KR are roll stiffnesses of the front and rear suspensions

KB are torsional stiffnesses of the body (much higher than KF, KR)

θ

( )2BPPT RL −=Equilibrium equations

RLaxle PPP +=

Torque T will reach a limit when right wheel lifts off, i.e., when PR = 0.

2maxBPT axle=

Maximum bump height Hmax that cause the right wheel to lift off theground is

T

axle

KBPH

2

2

max =

LP

RP

axleP

TAssume this axle isthe light axle.

B

Dynamic consideration

Increase the static moment by a factor ofa) 1.3 for road vehiclesb) 1.5-1.8 for off-road truck

Torsion bump height

Bump height, Hmax

Pawlowski Erz

0.2 m 0.2 m

≡

Torque Couple

This pure torsion load casecould not occur in practice.

However, it is importantbecause it generates verydifferent internal load instructure.

Longitudinal Load Case

1. Snap-clutch loads2. Accelerating/Braking3. Striking a bump

Maximum performance in longitudinal acceleration of a motor vehicleis determined by one of two limits– engine power or traction limits onthe drive wheel.

- At low speeds tire traction may be the limiting factor.- At high speeds engine power may account for the limits.

Accelerating

Traction-Limited Acceleration

⎟⎠⎞

⎜⎝⎛

dtdVMCG

Mg

La

h

FR RRFRµ

1) Front wheel drive

FRdtdVM µ=

MgRR RF =+

hdtdVMMgaLRR +=

Equilibrium equations

( )hL

haMgRR µµ

++

=

( )hLaLMgRF µ+

−=

Wheels reactions

CG

Mg

L

a

h

FR RR RRµ

⎟⎠⎞

⎜⎝⎛

dtdVM

2) Rear wheel drive

RRdtdVM µ=

MgRR RF =+

( ) hdtdVMaLMgLRF −−=

( )hL

ahLMgRF µµ

−−−

=

hLMgaRR µ−

=

Equilibrium equations Wheels reactions

Engine

Transmission

Driveshaft

Axle shaft&

wheels

Clutch

Differential

,eT

Nt ,It ,αe

Nf

eeI α,

ddI α,

cT

dT

aT

wwI α,

tη,

fη,

eT Engine torque

cT Torque input to transmission

dT Torque input to driveshaft

aT Torque on the axle

eI

tI

dI

wI

Rotational inertia of the engineRotational inertia of the transmissionRotational inertia of the driveshaft

Rotational inertia of the wheels andaxle shaft

eα Rotational acceleration of the enginedα Rotational acceleration of the driveshaft

wα Rotational acceleration of the wheel

ft NN , Gear ratio of the transmission and finaldrive

ft ηη , Power transmission efficiency of thetransmission and final drive

Power-Limited Acceleration

eeec ITT α−=

( ) ttetcd NITT ηα−=

( ) ffddda NITT ηα−=

The rotational accelerations are related by

wfd N αα =

wftdte NNN ααα ==

The flow of the torque from the engine to wheels can be derived as follows

Using the above equations, we can solve for the tractive force that canbe obtained from the engine as

( )[ ] 2222

raININNII

rNNT

F xwfdftte

ftftex +++−=

ηη

Effective inertia

Steady-state tractive forceavailable at the ground

Loss of tractive force due toinertia of the system.

wwax ITrF α−=

CG

Mg

L

a

h

FR RR

⎟⎠⎞

⎜⎝⎛

dtdVM

FRdtdVkM µ=⎟

⎠⎞

⎜⎝⎛ ( ) RR

dtdVMk µ=⎟

⎠⎞

⎜⎝⎛−1

( )L

haMgRRµ−

=

( )L

haLMgRFµ+−

=

Equilibrium equations Wheels reactions

RF RRdtdVM µµ +=

MgRR RF =+

hdtdVMMgaLRR −=

Braking

Striking a Bump

+H

2RP

PH

PV

θ θsinPPV =

θcosPPH =

⎟⎠⎞

⎜⎝⎛ −

=R

HRarcsinθ

(Assume the tire does not deflectexcessively)

Thusθtan

VH

PP =

Dynamic load factor = 4.5

Lateral loading

1. Cornering2. Overturning

Cornering

Sliding of tires can produce themaximum force of

latFCG

MgF µ=max Fmax

Overturning

latF

0→insideR MgRoutside →

h

latF

FP

RP

The worst possible condition occurswhen the vehicle is about to roll over.

B

KBMghFlat 2=Equilibrium condition

CG

CGL1

L2

Lateral force at

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=21

1

2 LLL

hMgBKPR

Rear wheel

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=21

2

2 LLL

hMgBKPF

Front wheel

K is dynamic safety factor.

Garrett suggested K = 1.75Others suggested K = 1.4

Tire Performance Curve

- Smaller increase in traction as thevertical load is increased.Tire

Verticalload

Traction

input output

- Cornering efficiency is defined as

loadVerticalTractionEffCornering =.

Ex.

Verticalload

Tractionavailable Cornering Eff.

5001,0001,5002,000

7001,0001,2501,500

1.401.000.830.75

Weight DistributionIn this section we will study the effect of weight distribution on acar handling behavior.

- Car weight 3000 lbs.- Front end weight 50 %- Left side weight bias 0 lbs- Load transfer from cornering 0 lbs

Example 1750 750

750 750

850 850

850 850

geffCornering 13.130003400. ==

- Car weight 3000 lbs.- Front end weight 50 %- Left side weight bias 0 lbs- Load transfer from cornering 1000 lbs

Example 2 750-500 750+500

750-500 750+500

450 1130

450 1130

geffCornering 05.130003160. ==

= 250

= 250

= 1250

= 1250

Cornering power decrease due to thelateral weight transfer.

- Car weight 3000 lbs.- Front end weight 60 %- Left side weight bias 0 lbs- Load transfer from cornering 1,000 lbs

Example 3 : Front-heavy car900-600 900+600

600-400 600+400

500 1250

380 1000

geffCorneringFront 03.118001850. ==

= 350

= 200

= 1500

= 1000

The car will tend to understeer whilecornering.

geffCorneringRear 27.112001520. ==

Terminology and Overview ofVehicle Structure Types

Basic Requirements

Purpose of structure

1. To maintain the shape of the vehicle2. To support the various loads applied to vehicle.

Design criteria (aim)1. To achieve sufficient levels of

strength and stiffness witha minimum mass.

2. To achieve acceptablecrash performance.

Strength

- Structure can lose its function by- Overstressing (yielding)- Buckling- joint failure- fatigue failure

Definition : Maximum force which the structure can withstand.

- Different load cases cause different local component loads.- The structure must have sufficient strength for all load cases.

Notes

Stiffness

1. Bending stiffness, KB

- Stiffness has an important influences on vehicle handling andvibrational behaviors and function of vehicle part such as doors.

The two stiffness definitions are

Relates the symmetrical vertical deflection of the point near thecenter of the wheelbase to multiples static loads on the vehicle.

2. Torsional stiffness, KT

Relates the angular deflection to an applied pure torque about thelongitudinal axis of the vehicle.

- Torsion case usually difficult to design for, so that the torsionalstiffness is often used as a benchmark of vehicle structure.

Satisfactory loadpaths.

Structure hassufficient stiffness

Satisfactory Structure

Most appropriatestructural type

for intendedapplication.

Correct layoutof structural

elements.

Good designof joint

Appropriate sizingof panels and

section

Satisfactory Structure

Correctly selectfailure mode

Correctly analyzecomponent loads

Modern structure typesTorsional constant, J

For thin wall closed sectionS

tAJ E24

=

AE is the enclosed areat is the wall thicknessS is distance around section perimeter

LGJKT =

Torsional stiffness, KT

G is material shear modulusL is length of member

Hence there is a great advantage in increasing the breadth and depthof the member.

Alpine Renault A310Lotus1. Backbone structure

Backbone chassisMade of triangulatedtubes

2. Triangulated tube structure

- Coachwork can consist of thin sheet metal cladding, attacheddirectly to the framework.

- add roll cage to passengercompartment).

- more increase in the torsionalstiffness.

3. Monocoque

4. Punt structure

Ford GT 40

- Floor member are of large closed section with good jointsbetween member.

- In many case the upper body is treated as structural insignificant.

Lotus Elise

5. Perimeter space frame

- Small section tubular members are built into ring-beam. - Each ring beam must be stiff locally at the corner.- Ring-beam are moderately effective at carrying local in-plane shear.

Introduction to“Simple Structural Surfaces” (SSS) Method

Definition

ta

b x

yz

My

Fx

Fz

3

121 tbI yy =

SSS is a plane structural element that can be considered as rigidonly in its own plane (i.e. flexible to out-of-plane load).

3

121 atI xx =

3

121 btI zz =

zzyy

xxyy

IIII

>>

>>

Examples of SSS

1. Panel

2. Swagedpanel

3. Panel with reinforcedhole

5. Pin jointedframework

4. Windscreenframe

Q1

Q2

Q1

Q2

Q1

Q2

Q1

Q2

Q1

Q2

Cornergussets

6. Sideframe

Examples of non-SSS

Panel-Boom (Stiffened shear web)

Fz

b

a

Panel(web)

Boom (flange)

The structure consists of a thin rectangularsheet to which a rod is bonded along eachedge.

Pin joint

- without a panel the structure is unstable.

- with a diagonal member the structure isstable statically determinate truss.

- with a panel, to make it a staticallydeterminate structure, we assume thatpanel carries only shear load.

Stiffener

Q1

Q2

K1

K2Fz

Q1

Q2

FBD.

Panel does not participate in producinginternal bending moment at the section.

Shear forceon the edge

Equilibrium equations

02 =−QFz

Top boom 011 =−KQBottom boom 021 =−KQ

Panel 021 =− aQbQ

zFQ =2Vertical boom

baF

baQQ z== 21

11 QK =

12 QK =

Q1

Q2

K1

K2Fz

Q1

Q2

Fz x

bxFz

bxFz

zFShear force

diagram

Bending momentdiagram

Shear force

zF

0

aFz

a

a x

x

Bending moment

Floor PanelFz

K3

K4w2

w1 l

Q4

Q4

Q3

Q3

Auxiliary beam

An auxiliary beam is added to carrythe vertical load Fz.

Equilibrium equations

0413 =− lQwQFloor panel

Cross beam 043 =−+ zFKK

( ) 02113 =−− wwFwK z

Shear forcediagram

Bending momentdiagram

Shear force

3K0

aFz

aa x

x

Bending moment

Simply supportend

4K

FzK1

K1

wa

b

K3

K2

A

B

C

D

E

Floor Panel : Edge Load

FzL

K2D

K3

E

FzCK1

K1

A

B

Cross beam

C

zCC EI

wF48

3

=δ

Longitudinal beam

( ) ( )222

6abL

baEIabF

L

zLL −−

+=δ

+

LC δδ =

zLFKKK =++ 3212

bKaK 32 =

Equilibrium & Compatibility equations

Deflection at loading point

( ) Cl

zLC

IbawbaIFIbaK 223

22

1 168

++=

( ) Cl

lzL

IbawbaIbIwFK 223

3

2 16++=

( ) Cl

lzL

IbawbaIaIwFK 223

3

2 16++=

Solve

Simple Box Structure

Bending Load Case4

5

3

2

6

1Rf Rf

K2

K2

K1

K1

K2

K3

Rr Rrw

h

abl

Fz

K2

K3K1

K1

K3

K3

f

r

02 1 =− zFK

022 2 =− KRf

1

5

022 3 =− KRr6

0132 =−+ KKK4

031 =− lKaK

Equilibrium equations

Note :Roof panel (No.3) carries no loads.

Pure Torsion Load Case4

5

3

2

6

1Rf’ Rf’

Q5

Q5

Q5

Rr’ Rr’w

h

abl

Q5

Q5

Q5

Q5

f

r

054 =−+′ wQhQfRf

046 =+ lQwQ

5

3

045 =− hQwQ6

4 065 =− hQlQ

Equilibrium equations

Q5

Q4

Q4

Q6

Q4

Q6

Q4Q6

Q6

Q4

Q4

Q6

Q6

Q6

Q6

Q4

Q4

0=′−′ rRfR rf

4

5

3

2

6

1Rf’ Rf’

Q5

Q5

Q5

Rr’ Rr’

Q5

Q5

Q5

Q5

f

r

Q5

Q4

Q4

Q6

Q4

Q6

Q4Q6

Q6

Q4

Q4

Q6

Q6

Q6

Q6

Q4

Q4

Missing Roof Panel : Torsion Load Case

Q4 0 Q6 0No. 3

Q5 0No. 2, 4

No. 6 0→′rR No. 5 0→′rR

Cannot carry a torsion load

Represent Vehicle Structures by SSS(Pawlowski 1964)

Typical Car Bodywith SSS Idealization

Saloon Car

12

3Central longitudinal tunnel

Floor panel Front seat cross beam

7Engine rail

8Dash panel

13Windscreen

Cowl12

Side panel11

Rear quarter panel10

4 5

9

161514

Luggagecompartment floor

Backlightframe

Rear seat cross beam

SSS 1, 2, 4 : Carry seat loads, support SSS7SSS 6 : Carry luggage loads, rear suspension loads

6

SSS 7 : Carry engine/transmission loads, front suspension loadsSSS 8 : Support SSS7SSS 9 : Transfer load to SSS10

RFRR

SSS 3, 8, 12-16, 9, 5, 4, 11 : Shear carrying member

Station Wagon

1 23

5

4

6

RF

RR

7

8

9

10

11

1213

14

15

SSS 7 : Carry rear suspension loadSSS 10 : Carry front suspension loadSSS 8, 9 : Support SSS10SSS 11, 2 : Support SSS8

A-pillar

SSS 4, 6, 7, 11-15 : Shear carrying members

Van

RF

RF

RR

RR

12

3

4

5

6

7

89

10

SSS 1-6 : Carry bending loadSSS 5-10 : Carry torsion load

Introduction to Vehicle Structure Preliminary Design

Synthesis vs Analysis

Given : Beam with a combination of uniformly distributed load andconcentrated load.

What section size is required to support these loads ?Question

Given : Beam with a combination of uniformly distributed load andconcentrated load. Also beam type, length, cross-section shape,size and material used are given.

Can this structure carry the load ?Question

1. Solution Shape, dimensions, material etc.2. Procedure to Optimization

achieve a solution

Synthesis

Analysis

Suggested Steps1. Estimate the loads and loading conditions

(it is recommended to start first with bending and torsionload case)

2. Draw FBD and loading diagram3. Formulate a system of equation to solve for the edge forces.4. Construct a shear and bending moment diagram.

Step 1 to 4 should be repeated for local subunits.1. Lower structure2. Dash and rear seat panel3. A-pillar4. B,C,D pillars.5. Cantrails, windshield, and backlight glass.

Requirement : Two different size vehicles with the same platform.

Initial decision1. Make a floor panel wider and longer.2. Sill cross-section is the same for both sizes.3. Motor compartment side panel structure is the same for both sizes.4. Width of upper and lower front cross member is different.

Possible solution1. The frame member sections must be designed so that the stiffness of

the larger vehicle is not compromised. But, the smaller vehicle isallowed to be overdesigned.

2. The frame members are designed based on the smaller vehicle. Anydifferences in the larger vehicle are to be solved by additionalreinforcements which must be compatible with the manufacturingprocess.

Example : Application at starting point

Design Guidelines1. SSS can resist only tension, compression and shear forces in its own

plane.2. Stiffener (integral or add-on) are required to improve the capability of

compression load carrying.3. Stiffeners are required in order to resist small, distributed load normal

to the SSS4. Large concentrated loads must be resisted by transmitting loads to the

plane of an adjacent SSS or use a stiffening member to distribute theload.

Suspension reacting support

Suspension load

Rearlongitudinal rail

Rear compartment panel

Use bulkhead Move load application Transfer rail load to anSSS in-plane

Alternative design

Steering column supportDash panel

Lowersupport

Uppersupport

Steering column assembly performance criteria require are

1. Meet a minimum natural frequency target to assure vibration isolation from road and engine idle excitation. Minimize wheel deflection

2. Accommodate occupant safety and vehicle crashworthiness objectives.

Vertical deflection at the wheel is ( )EI

ALWAY3

2 += Reduce W, A, L+A

Increase EI

FL

FU

AL

W

Y

FL

FU

AL

W

FBD : Steering columnY

FL

FU

( )L

ALWFU+

=

LWAFL =

FdFd

WFd =

( ) ( )H

BL

WABLL

ALW

Fx

−++

=

Dash panel Support bracketB

Fx

Fx

H

FBD : Support bracket

Force analysis

Construction

Fx

Fx

Fd

Fx

Fd

Cowl air plenum panelCowl bar beam

Fx

Fd

Fx FxR

R

Front bodyhinge pillar

Transverse beam

Alternative design 1 Alternative design 2 Alternative design 3

Need locally reinforcement